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题目大意:经典的埃及分数问题。
代码如下:
# include<iostream> # include<cstdio> # include<cstring> # include<algorithm> using namespace std; # define LL long long int num[5],a,b,k; LL ans[10000],v[10000]; LL gcd(LL a,LL b) { return (b==0)?a:gcd(b,a%b); } int get_first(LL a,LL b) { return b/a+1; } bool judge(int val) { for(int i=0;i<k;++i) if(val==num[i]) return false; return true; } bool better(int d) { for(int i=d;i>=0;--i) if(ans[i]!=v[i]) return ans[i]==-1||v[i]<ans[i]; return false; } bool dfs(int cur,int maxd,int from,LL aa,LL bb) { if(cur==maxd){ if(bb%aa) return false; v[cur]=bb/aa; if(!judge(v[cur])) return false; if(better(cur)){ for(int i=0;i<=cur;++i) ans[i]=v[i]; } return true; } bool ok=false; from=max(from,get_first(aa,bb)); for(int i=from;;++i){ while(!judge(i)) ++i; if(bb*(maxd-cur+1)<=i*aa) break; v[cur]=i; LL b2=bb*i; LL a2=aa*i-bb; LL g=gcd(a2,b2); if(dfs(cur+1,maxd,i+1,a2/g,b2/g)) ok=true; } return ok; } void solve() { for(int maxd=1;;++maxd){ memset(ans,-1,sizeof(ans)); if(dfs(0,maxd,get_first(a,b),a,b)){ printf("%d/%d=",a,b); for(int i=0;i<=maxd;++i) printf("1/%lld%c",ans[i],(i==maxd)?‘\n‘:‘+‘); break; } } } int main() { int T,cas=0; scanf("%d",&T); while(T--) { scanf("%d%d%d",&a,&b,&k); for(int i=0;i<k;++i) scanf("%d",num+i); printf("Case %d: ",++cas); solve(); } return 0; }
UVA-12558 Egyptian Fractions (HARD version) (IDA* 或 迭代加深搜索)
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原文地址:http://www.cnblogs.com/20143605--pcx/p/4824313.html